Chemistry
Chemical Quantities
Lesson 6
Lesson Plan
David V. Fansler
The Mole: A Measurement of Matter
Objectives: Describe how Avogadro’s number is related to a mole of any substance;
Calculate the mass of a mole of any substance
- What is a Mole?
o We live in a quantitative world – numbers describe everything
around us.
Grade on last test
Amount of money in our pockets
How many times you heard your favorite song on the radio
yesterday
How far is it from your house to the school
How much do you weigh
o In chemistry, chemist ask questions that also involve numbers
How many kg of iron can be obtained from a kg of iron ore
How many grams of hydrogen and nitrogen must be
combined to make 200g of the fertilizer ammonia (NH3)
o As we move forward in our chemistry, we will be involved in
analyzing the composition of samples of matter
Including calculations relating quantities of reactants and
products to chemical equations
To do this we must be able to measure matter
o How do you measure matter?
Count the number of CD’s in your collection
Count the number of stuffed bears in your collection
How do you buy beans? Buy counting how many you are
buying? Typically by weight.
Gold is sold by weight
Potatoes are sold by weight
How do you buy milk? By volume
Gasoline is by volume
Shampoo is by volume
There are numerous ways to measure matter
o Common units of measurement
Socks are sold in ____ (pairs) meaning 2
Shoes are sold in pairs
Eggs are sold by the dozen – 12
• A dozen of anything is 12
• Except a baker’s dozen which is 13
Apples are sold in 3 different ways
• By count (meaning 5)
David V. Fansler – Beddingfield High School – Page 1
Chemistry Lesson #6 - Chemical Quantities
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By weight ($2/pound)
Or by the volume – bushel ($12/bushel)
We could set up an equivalency of count, mass and
volume
o By count: 1 dozen apples = 12 apples
o By mass: 1 dozen apples = 2.0 kg apples
o By volume: 1 dozen apples = 0.20 bushel
apples
o The problem in measuring atoms or molecules is they are too small
to be seen and too light to be weighed individually.
o The Mole (mol) – the mole is like a dozen – where a dozen is 12 of
anything, a mole is 6.02 x 1023 of anything’s.
A mole of apples would be 6.02 x 1023 apples
A mole of doughnuts would be 6.02 x 1023 doughnuts
6.02 x 1023 is called Avogadro’s number
• In honor of Amedo Avogadro di Quadrenga
• He did not determine the number, rather he clarified
the difference between atoms and molecules
o Avogadro’s number is used with the term “representative particles”
This could be atoms, molecules or formula units
The representative particle for most elements is the atom
There are 7 elements that normally occur as diatomic
molecules (H2, N2, O2, F2, Cl2, Br2, and I2 )
• The representative particle for these elements and
all molecular compounds is the molecule
• H2O, or SO2
The representative particle for ionic compounds is the
formula unit
• NaCl (1:1) or CaCl2 (1:2)
Substance
Atomic nitrogen
Nitrogen gas
Water
Calcium ion
Calcium fluoride
Sucrose
Representative Particles and Moles
Representative
Chemical
Representative particles
Particle
Formula
in 1.00 mol
Atom
N
6.02 x 1023
Molecule
N2
6.02 x 1023
6.02 x 1023
Molecule
H2O
6.02 x 1023
Ca2+
Ion
CaF2
6.02 x 1023
Formula Unit
C12H22O11
Molecule
6.02 x 1023
Example Problem #1
How many moles of magnesium is 1.25 x 1023 atoms of magnesium?
Known
Unknown
23
Number of atoms = 1.25 x 10 atoms of Mg
moles = ? mol Mg
1 mol Mg = 6.02 x 1023 atoms of Mg
David V. Fansler – Beddingfield High School – Page 2
Chemistry Lesson #6 - Chemical Quantities
The desired conversion is atoms
moles
1 mol Mg
The conversion factor is
6.02 x 1023 atoms Mg
Multiplying the known number of atoms of Mg by the conversion yields:
1 mol Mg
1.25 x 1023 atoms Mg •
6.02 x 1023 atoms Mg
1 mol Mg
= .208 mol Mg =
1.25 x 1023 atoms Mg •
6.02 x 1023 atoms Mg
= 2.08 x 10−1 mol Mg
Does the answer make sense?
Because the given number of atoms is less than ¼ of Avogadro’s number,
the answer should be less than ¼ mole of atoms. The answer should have
three significant digits.
-
Determining the number of atoms in a mole of a compound
o First you must know how many atoms are in a representative
particle of the compound – this is determined by the chemical
formula
CO2 has three atoms (1 carbon and 2 oxygen)
A mole of carbon dioxide contains Avogadro’s number of
CO2 molecules
Thus a mole of carbon dioxide contains three times
Avogadro’s number of atoms
If we were talking about CO, then since it consists of two
atom per molecule, there would be twice Avogadro’s
number of atoms in 1 mole
Sample Problem #2
How many atoms are in 2.12 mol of propane (C3H8)?
Known
Unknown
Number of atoms = ?
Number of moles = 2.12 mol (C3H8)?
atoms
1 mol C3H8 = 6.02 x 1023 molecules C3H8
1 molecule C3H8 = 11 atoms
(3 carbon and 8 hydrogen)
The desired conversion is moles
molecules
atoms
23
6.02 x 10 molecules C3 H 8
The first conversion factor is
1 mol C3 H 8
11 atoms
The second conversion factor is
1 molecule C3 H 8
6.02 x 1023 molecules C3 H 8
11 atoms
x
1 molecule C3 H 8
1 mol C3 H 8
23
25
= 140.39 x 10 atoms = 1.40 x 10 atoms
So 2.12 mol C3 H 8 x
David V. Fansler – Beddingfield High School – Page 3
Chemistry Lesson #6 - Chemical Quantities
Does the result make sense?
Because there are 11 atoms in each propane molecule and more
than 2 molecules of propane, then the answer should be at least
2*10 times more than Avogadro’s number . The answer has 3
significant digits based on the given measurements
Practice Problems:
1. How many atoms are in 1.00 mole of sucrose (C12H22O11)?
6.02 x 1023 molecules C12 H 22O11
45 atoms
1.00 mol C12 H 22O11 x
=
x
1 molecule C12 H 22O11
1 mol C12 H 22O11
270.90 x 1023 atoms = 2.71 x 1025 atoms
2. How many atoms of C are in 2.0 moles of C12H22O11?
6.02 x 1023 molecules C12 H 22O11
12 atoms
2.00 mol C12 H 22O11 x
x
=
1 mol C12 H 22O11
1 molecule C12 H 22O11
144.48 x 1023 atoms = 1.44 x 1025 atoms
3. How many atoms of O are in 2.0 moles of C12H22O11?
6.02 x 1023 molecules C12 H 22O11
11 atoms
3.65 mol C12 H 22O11 x
=
x
1 molecule C12 H 22O11
1 mol C12 H 22O11
241.70 x 1023 atoms = 2.42 x 1025 atoms
4.
How many atoms of H are in 2.0 moles of C12H22O11?
6.02 x 1023 molecules C12 H 22O11
22 atoms
2.00 mol C12 H 22O11 x
=
x
1 molecule C12 H 22O11
1 mol C12 H 22O11
264.88 x 1023 atoms = 2.65 x 1025 atoms
5. How many atoms are there in 1.14 mol of SO3?
6.02 x 1023 molecules SO3
4 atoms
1.14 mol SO3 x
=
x
1 molecule SO3
1 mol SO3
27.45 x 1023 atoms = 2.75 x 1024 atoms
6. How many moles are there in 4.65 x 1024 molecules of NO2?
1 mol NO2
4.65 x 1024 molecules NO2 x
= .7724 x 101 mol = 7.72 mol
6.02 x 1023 molecules NO2
-
How big is Avogadro’s number?
o Use the animal mole for a basis
Typical mole is 15 cm long, 5 cm high, 150g mass
150 g
6.02 x 1023 animal-mole x
=9.03 x 1022 kg
1 animal − mole
David V. Fansler – Beddingfield High School – Page 4
Chemistry Lesson #6 - Chemical Quantities
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more than 1% the mass of the Earth
1.3 times the mass of the moon
60 times the mass of the Earth’s oceans
If spread over the Earth, it would create a layer 8
million animals thick
15 cm
6.02 x 1023 animal-mole x
=9.03 x 1019 km
1 animal − mole
• long enough that if lined up end to end the animals
would stretch from Earth to the nearest star, Alpha
Centauri and back to earth again
o If you make 6 billion stacks of animals and set one person to counting
one stack at the rate of 1000 animals per second, it would take more
than 3000 years to count 6.02 x 1023 animals.
o Avogadro’s number is really big!
The Mass of a Mole of an element
o Working with individual atoms, or even a billion atoms would be
impossible since there would be too little of the substance to
handle.
o Chemists get around this by working with grams of a substance
o By using the atomic mass of an element, expressed in grams, we
have the gram atomic mass
The gam of carbon is 12.0 g
The gam of Hydrogen is 1.0 g
The gam of Oxygen is 16.0 g
What is the gam of Hg and Fe? (200.5 g, 55.8 g)
o The gram atomic mass of any element contains 6.02 x 1023 atoms
of that element
o Remember that the atomic mass unit was based on carbon-12 and
the there were 12 amu in one cabon-12 atom?
And that hydrogen atom had 1 amu?
Then just as 1 atom of carbon weighs 12 times 1 atom of
hydrogen, the 100 atoms of carbon weighs 12 times 100
atoms of hydrogen.
In the same way, any number of carbon atoms will weigh
12 times the same number of hydrogen atoms.
Or if we were to compare carbon and oxygen, we would
find that 12 g of carbon atoms would weigh as much as 16
g of oxygen atoms – and both would contain the same
number of atoms – Avogadro’s number of atoms.
o So now the mole can be defined as the amount of substance that
contains as many representative particles as the number of atoms in
12 g of carbon-12
12 g of carbon is 1 mole of carbon and contains 6.02 x 1023
atoms
1 g of hydrogen is 1 mole of hydrogen and contains 6.02 x
1023 atoms
David V. Fansler – Beddingfield High School – Page 5
Chemistry Lesson #6 - Chemical Quantities
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16 g of oxygen is 1 mole of oxygen and contains 6.02 x
1023 atoms
The gram atomic mass is the mass of 1 mole of atoms of
any element
The Mass of a Mole of a compound
o Like a atom, a molecule is far too small to be weighed, so again we
use the mole
o First we must know the formula of the compound – this tells us the
number of atoms of each element in a representative particle.
o
SO3 – has 1 sulfur and 3 oxygen atoms
Add the atomic mass of all the atoms making up one
molecule (1 * 32.1 amu + 3 * 16.0 amu = 32.1 amu + 48
amu = 80.1 amu (the molecular mass of sulfur trioxide)
When working with atoms, we used the gram atomic mass,
with compounds we use gram molecular mass – same
concept, but for compounds rather than atoms. That is 1
gmm of a compound is 1 mol of that compound
1 amu = 1gmm
So 80.1 amu = 80.1 gmm = 1mol of SO3
Examples:
1. What is the gram molecular mass of hydrogen peroxide (H2O2)?
Known
Unknown
gmm = ?g
Molecular formula = H2O2
1 gam H = 1 mol H = 1.0 g H
1 gam O = 1 mol O = 16.0 g O
Since we know that there is 2 mol of H and 2 mol of O in 1 mol of
hydrogen peroxide, we must convert moles of atoms into grams by using
the conversion factors (g/mol) based on the gram atomic mass of each
element. The sum of the masses of the elements gives the gram molecular
weight.
1.0 g H
2 mol H x
= 2.0 g H
1 mol H
16.0 g O
2 mol O x
= 32.0 g O
1 mol O
2.0 g H + 32.0 g O = 34.0 g = gmm of H2O2
2.
1 mol of glucose molecules (blood sugar) C6H12O6
6 carbons, 12 hydrogens, and 6 oxygens
6 * 12.0 + 12 * 1.0+6 * 16.0 = 180.0 g C6H12O6 = 1 gmm C6H12O6
David V. Fansler – Beddingfield High School – Page 6
Chemistry Lesson #6 - Chemical Quantities
3.
1 mol of H2O (water)
2 hydrogen and 1 oxygen
2 * 1.0 + 1 * 16.0 = 18.0 g H2O = 1 gmm H2O
4.
1 mol of paradichlorobenzene molecules (moth crystals) C6H4Cl2
6 carbons, 4 hydrogens, and 2 chlorines
6 * 12.0 + 4 * 1.00 + 2 * 35.5 = 147.0 g C6H4Cl2 = 1 gmm C6H4Cl2
5.
What is the gram molecular mass of C2H6?
2 carbons and 6 hydrogens
2 * 12.0 + 6 * 1.0 = 30.0 g
6.
What is the gram molecular mass of PCl3?
1 phosphorous and 3 chlorine atoms
1 * 31.0 + 3 * 35.5 = 137.5 g
7.
What is the gram molecular mass of C3H7OH?
3 carbon, 8 hydrogen, and 1oxygen
3 * 12.0 + 8 * 1.0 + 1 * 16.0 = 60.0 g
8.
What is the gram molecular mass of N2O5?
2 nitrogen and 5 oxygen
2 * 14.0 + 5 * 16.0 = 108.0 g
9.
What is the mass of 1.00 mole of chlorine?
Cl occurs as a diatomic atom, so it is Cl2
2* 35.5 = 71.0g
10.
What is the mass of 1.00 mole of nitrogen dioxide?
NO2
1 nitrogen and 2 chlorine
1 * 14.0 + 2 * 16.0 = 46.0 g
11.
What is the mass of 1.00 mole of carbon tetrabromide?
CBr4
1 carbon and 4 bromine
1 * 12.0 + 4 * 79.9 = 331.6 g
12.
What is the mass of 1.00 mole of silicon dioxide?
SiO2
1 silicon and 2 oxygen
1 * 28.1 + 2 * 16.0 = 60.1 g
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How about ionic compounds?
o Remember the representative unit for a molecular compound is the
molecule, but for an ionic compound it is the formula unit
David V. Fansler – Beddingfield High School – Page 7
Chemistry Lesson #6 - Chemical Quantities
o When calculating the mass of a mole of an ionic compound we use
the gram formula mass.
o gfm is just like the gmm, in that the gfm equals te formula mass
expressed in grams – simply sum the atomic masses of the ions in
the formula unit of the compound
Examples:
1.
What is the gram formula mass of ammonium carbonate ((NH4)2CO3)?
Knowns
Unknowns
gfm = ? g
Formula unit = (NO4)2CO3
1 gam N = 1 mol N – 14.0g N
1 gam H = 1 mol H = 1.0 g H
1 gam C = 1 mol C = 12.0 g C
1 gam O = 1 mol O = 16.0 g O
The formula shows that a mole of this ionic compound is composed of 2
mol of nitrogen, 8 mol of hydrogen, 1 mol of carbon and 3 mol of oxygen.
Moles of atoms are converted to grams using the conversion factor based
on the gram atomic masses. The sum of the masses of the elements gives
the gram formula mass.
14.0 g N
= 28.0 g N
2 mol N x
1 mol N
1.0 g H
= 8.0 g H
8 mol N x
1 mol H
12.0 g C
= 12.0 g C
1 mol C x
1 mol C
16.0 g O
= 48.0 g O
3 mol N x
1 mol O
so 28.0 g N + 8.0 g H + 12.0 g C + 48.0 g O = 96.0 g as the gram formula
mass of (NO4)2CO3
2.
Calculate the gram formula mass of K2O.
2 potassium, 1 oxide
2 * 39.1 g + 1 * 16.0 g O = 94.2 g
3.
Calculate the gram formula mass of CaSO4.
1 calcium, 1 sulfur and 4 oxygen
1 * 40.1 g + 1 * 32.1 g + 4 * 16.0 g = 136.2 g
4.
Calculate the gram formula mass of CuI2.
1 copper and 2 iodide
1 * 63.6 g + 2 * 126.9 g = 317.4 g
5.
Find the gram formula mass of barium fluoride.
BaF2 = 1 barium and 2 fluoride
David V. Fansler – Beddingfield High School – Page 8
Chemistry Lesson #6 - Chemical Quantities
1 * 137.3 g + 2 * 19.0 g = 175.3 g
6.
Find the gram formula mass of strontium cyanide.
Sr(CN)2 = 1 strontium, 2 carbon and 2 nitrogen
1 * 87.6 g + 2 * 12.0 g + 2 * 14.0 g = 139.6 g
6.
Find the gram formula mass of sodium hydrogen carbonate.
NaHCO3 = 1 sodium, 1 hydrogen, 1 carbon, and 3 oxygen
1 * 23.0 g + 1 * 1.0 g + 1 * 12.0 g + 3 * 16.0 g = 84.0 g
7.
Find the gram formula mass of aluminum sulfite.
Al2(SO3)3 = 2 aluminum, 3 sulfur and 9 oxygen
2 * 27.0 g + 3 * 32.1 g + 9 * 16.0 g = 294.3 g
Section Review Problems
Mole-Mass and Mole-Volume Relationships
Objectives: Use the molar mass to convert between mass and moles of a substance; Use
the mole to convert among measurements of mass, volume, and number of particles
- The Molar Mass of a Substance
o Just learned three new terms
Gram atomic mass (for elements)
Gram molecular mass (for molecular compounds)
Gram formula mass (for ionic compounds)
Each represents a mole for a particular substance
- To represent all of the above we use molar mass to represent the mass to
make one mole of any substance.
o This can cause problems – for example if we talk about a molar
mass of oxygen, are we talking about atoms of oxygen (O), then
the molar mass is 16.0 g. If we mean the oxygen molecule (O2),
then the molar mass is 32.0 g (2 x 16.0 g).
Chemist use more specific terms when questions may be
raised.
Sample Problems:
1.
How many grams are in 9.45 mol of dinitrogen trioxide (N2O3)?
Known
Unknown
mass = ? g N2O3
9.45 mol N2O3
1 mol N2O3 = 76.0 g
moles
grams
76.0 g N 2O3
= 718.2 g N 2O3 = 718 g N 2O3
9.45 mol N 2O3 x
1.00 mol N 2O3
Does the result make sense?
Since 1 mole = 76 g, then we have just under 10 mol so the answer
should be about 700. Be sure to round your answer off to the proper
number of significant digits.
David V. Fansler – Beddingfield High School – Page 9
Chemistry Lesson #6 - Chemical Quantities
2.
3.
4.
5.
6.
7.
8.
Find the mass, in grams of 3.32 mol K.
39.1 g K
= 129.8 g K = 130 g K
3.32 mol K x
1 mol K
Find the mass, in grams of 4.52 x 10-3 mol C20H42.
282.0 g C20 H 42
= 1.274 g C20 H 42 = 1.27 g C20 H 42
4.52 x 10−3 mol C20 H 42 =
1 mol C20 H 42
Find the mass, in grams of 0.0112 mol K2CO3.
138.2 g K 2CO3
= 1.547 g K 2CO3 = 1.55 g K 2CO3
0.0112 mol K 2CO3 =
1 mol K 2CO3
Calculate the mass, in grams of 2.50 mol sodium sulfate.
Sodium Sulfate is Na2SO4
142.1 g Na2 SO4
= 355.25 g Na2 SO4 = 355 g Na2 SO4
2.50 mol Na2 SO4 =
1 mol Na2 SO4
Calculate the mass, in grams of 2.50 mol iron(II) hydroxide.
Iron(II) hydroxide is Fe(OH)2
89.9 g Fe(OH ) 2
= 224.75 g Fe(OH ) 2 = 225 g Fe(OH ) 2
2.50 mol Fe(OH ) 2 =
1 mol Fe(OH ) 2
Find the number of moles in 92.2 g of iron (III) oxide (Fe2 O3).
Known
Unknown
Number of moles = ? mol Fe2 O3
Mass – 92.2 g Fe2 O3
1 mol Fe2 O3 = 159.6 Fe2 O3
conversion is grams
moles
1.00 mol Fe2O3
= 0.5776 mol Fe2O3 = 0.578 mol Fe2O3
92.2 g Fe2O3 x
159.6 g Fe2O3
Does the result make sense?
Because the mass (about 90 g) is slightly larger than the mass of ½ mol of
Fe2O3 (about 160 g), the answer should be slightly larger than ½ mole.
Find the number of moles in 3.70 x 10-1 g B.
1.00 mol B 3.70 x 10−1 mol B
=
= 3.426 x 10−2 mol B = 3.43 x 10−2 mol B
3.70 x 10−1 g B x
10.8 g B
10.8
Find the number of moles in 2.74 g TiO2.
1.00 mol TiO2 2.74 mol TiO2
=
= 3.429 x 10−2 mol TiO2 = 3.45 x 10−2 mol TiO2
2.74 g TiO2 x
79.9 g TiO2
79.9
10.
Find the number of moles in 847 g (NH4)2CO3.
1.00 mol ( NH 4 ) 2 CO3 847 mol ( NH 4 ) 2 CO3
=
=
847 g ( NH 4 ) 2 CO3 x
96.0 g ( NH 4 ) 2 CO3
96.0
9.
10.
8.822 mol ( NH 4 ) 2 CO3 = 8.82 mol ( NH 4 ) 2 CO3
Calculate the number of moles in 75.0 g of dinitrogen trioxide (N2O3).
David V. Fansler – Beddingfield High School – Page 10
Chemistry Lesson #6 - Chemical Quantities
1 mol N 2O3
= .9868 mol N 2O3 = 9.87 x 10−1 mol N 2O3
76.0 g N 2O3
Calculate the number of moles in 75.0 g of nitrogen gas (N2).
75.0 g N 2O3 x
11.
1 mol N 2
= 2.678 mol N 2 = 2.68 mol N 2
28.0 g N 2
12.
Calculate the number of moles in 75.0 g of sodium oxide (Na2O).
1 mol N 2O
75.0 g Na2O x
= 1.209 mol Na2O = 1.21 mol Na2O
62.0 g N 2O
- The Volume of a Mole of Gas
o From examples we did earlier, we found that 1 mol of glucose had
a gmm of 180.0 g, 1 mol of water had a gmm of 18.0 g and 1mole
of paradichlorobenzene had a gmm of 147.0 g. The volume that
these compounds occupy vary according to density.
o Gases are more predictable under the same physical conditions.
o When talking about physical characteristics of the states of matter
(solid, liquid and gas), that a gas expanded with heat – so the
volume of a gas will change with an increase of heat. If a gas is in
a restricted volume, then heating the gas will raise the pressure of
the gas.
Because of this variation, the volume of a gas is usually
measured at a standard temperature and pressure –
abbreviated as STP.
Standard temperature is 0ºC
Standard pressure is 101.3 kPa, or 1 atmosphere (atm)
o At STP, 1 mol of gas occupies a volume of 22.4 L
o 22.4 L is known as the molar volume of a gas
o Because 1 mol of anything contains Avogadro’s number of
representative particles, then 22.4 L of gas at STP contains 6.02 x
1023 representative particles of that gas
Would this differ for gaseous elements compared with
gaseous compounds? (no – a mole one gas contains just as
many representative particles as another gas – think of
molecules and atoms)
Would 22.4 L of one gas have the same mass as 22.4 L of
another gas at STP?
• Probably not – since gases are composed of
different elements, just as solid compounds are
composed different elements and have different
mass
• Only gasses with the same molar mass at STP
would have equal masses for equal volumes at STP
Sample Problems
1.
Determine the volume, in liters, of 0.60 mol SO2 gas at STP.
75.0 g N 2 x
David V. Fansler – Beddingfield High School – Page 11
Chemistry Lesson #6 - Chemical Quantities
2.
Known
Unknown
volume = ? L SO2
0.60 mol SO2
1 mol SO2 = 22.4 L SO2
Conversion is moles
liters
22.4 L SO2
0.60 mol SO2 x
= 13.44 L SO2 = 13 L SO2
1 mol SO2
What is the volume at STP of 3.20 x 10-3 mol CO2?
22.4 L CO 2
= 71.68 x 10−3 L CO2 = 7.17 x 10−2 L CO 2
3.20 x 10-3 mol CO 2 x
1 mol CO 2
What is the volume at STP of 0.960 mol CH4?
22.4 L CH 4
0.960 mol CH 4 x
= 21.50 L CH 4 = 21.5 L CH 4
1 mol CH 4
4.
What is the volume at STP of 3.70 mol N2?
22.4 L N 2
3.70 mol N 2 x
= 82.88 L N 2 = 8.29 L N 2
1 mol N 2
5.
Assuming STP, how many moles are in 67.2 L SO4?
1 mol SO 2
67.2 L SO 2 x
= 3.00 L SO 2
22.4 L SO 2
6.
Assuming STP, how many moles are in 0.880 L He?
1 mol He
= 3.928 x 10-2 L He = 3.93 x 10-2 L He
0.880 L He x
22.4 L He
7.
Assuming STP, how many moles are in 1.00 x 103 L C2H6?
1 mol C2 H 6
1.00 x 103 L C 2 H 6 x
= 44.64 L C2 H 6 = 4.46 x 101 L C2 H 6
22.4 L C2 H 6
mass
which for our lab was
- Remember density? For a solid it was the
volume
measured in g/ml. For gases, density is usually measured in g/L
o The experimentally determined density of a gas at STP is sued to
calculate the molar mass of that gas
The gas can be an element or a compound
What makes a balloon float or sink?
• Density of gas in the balloon compared to the
surrounding air
o Balloons filled with by mouth contain the
same air as they are surrounded by plus the
weight of the balloon
o Balloons filled with helium tend to float in
air as long as the density difference is great
enough to include the weight of the balloon
o Would a helium balloon float in an
atmosphere of helium? (no)
Example Problems:
3.
David V. Fansler – Beddingfield High School – Page 12
Chemistry Lesson #6 - Chemical Quantities
1.
The density of a gaseous compound containing carbon and oxygen is
1.964 g/L at STP. Determine the molar mass of the compound.
Known
Unknown
Density = 1.964 g/L
molar mass = ? g/mol
1 mol (gas at STP) = 22.4 L
g
g
Conversion →
L
mol
1.964 g 22.4 L
x
= 43.99 g / mol = 44.0 g / mol
L
1 mol
Does the result make sense?
The ratio of the calculated mass to the volume is about 2, which is about
what the density was.
2.
A gaseous compound composed of sulfur and oxygen that is linked to the
formation of acid rain has a density of 3.58 g/L at STP. What is the molar
mass of this gas?
3.58 g 22.4 L
x
= 80.192 g / mol = 80.2 g / mol
L
1 mol
What is the density of krypton gas at STP?
83.8 g 1 mol
x
= 3.741 g / mol = 3.74 g / mol
1 mol 22.4 L
The Mole Road Map
o We have now examined the mole in terms of particles, mass, and
volume of gases at STP. The below figure shows the relationship
of these terms – note that we must pass through the mole as an
intermediate step when converting from one to another.
3.
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David V. Fansler – Beddingfield High School – Page 13
Chemistry Lesson #6 - Chemical Quantities
Section Problems 24 – 28 p 186
Percent Composition and Chemical Formulas
Objectives: calculate the percent composition of a substance from its chemical formula
or experimental data; Derive the empirical formula and the molecular formula of a
compound from experimental data.
-
Calculating the Percent Composition of a Compound
o When taking care of a yard, it is important to feed the yard what it
needs to grow – typically this would be a mixture of nitrogen,
phosphorus and potassium. Most fertilizers are labeled with 3
numbers, such as 15-10-15, indicating the amount of each element
in the mixture.
o The relative amount of each element in a compound is expressed in
percent composition or percent mass
o The percent composition or percent mass of a compound will
contain as many values as there are elements in a compound, and
they must total to 100.0%
grams of element E
o % mass of element E =
x 100%
grams of compound
Example
If an 8.20 g of magnesium combines completely with 5.40 g of oxygen to form a
compound, what is the percent composition of this compound?
Known
Unknown
Mass of magnesium = 8.20 g
Percent Mg = ?% Mg
Mass of oxygen = 5.40 g
Percent O = ?% O
Mass of compound = 8.20 g + 5.40 g = 13.60 g
% Mg =
mass of Mg
x 100%
grams of compound
8.2 g
x 100% = 60.29% = 60.3%
13.6 g
mass of O
%O =
x 100%
Grams of compound
% Mg =
%O =
5.40 g
x 100% = 39.70% = 39.7%
13.60 g
Checking our results: 60.3% + 39.7% = 100.0%
Practice Problems
If 9.03 g of Mg combines completely with 3.48 g N to form a compound, what is
the percent composition?
9.03 g + 3.48 g = 12.51 g total mass of compound
David V. Fansler – Beddingfield High School – Page 14
Chemistry Lesson #6 - Chemical Quantities
% Mg =
9.03 g
x 100% = 72.2%
12.51 g
3.48 g
x 100% = 27.8%
12.51 g
If 29.0 g of Ag combines completely with 4.30 g S to from a compound, then
what is the percent composition?
29.0 g
% Ag =
x 100% = 87.1%
33.30 g
%N =
4.30 g
x 100% = 12.9%
33.30 g
When a 14.2 g sample of mercury(II) oxide is decomposed into its elements by
heating, 13.2 g Hg is obtained. What is the percent composition of this of the
compound?
13.2 g
% Hg =
x 100% = 93.0%
14.2 g
%S=
%O =
-
1.0 g
x 100% = 7.0%
14.2 g
So far we have calculated the percent by mass of an element in a
compound – now we learn to calculate the % composition of a compound
o With % mass we used the values of a known amount of a
compound and the known amount of the substances in that
compound
o With % composition, we are not concerned with being given a
known value of a compound, rather we use the molar mass of the
compound and the grams of an element in mol of the compound.
grams of element in 1 mol compound
x 100%
% mass =
molar mass of compound
o Compare two compounds with the same elements, but with
different chemical formulas and how the percent composition
differs
David V. Fansler – Beddingfield High School – Page 15
Chemistry Lesson #6 - Chemical Quantities
Sample Problems
Calculate the percent composition of propane (C3H8)
Known
Unknown
percent C = ?%C
molar mass of C3H8 = 44.0g/mol
percent H = ?%H
Mass of C in 1 mol C3H8 = 36.0 g
Mass of H in 1 mol C3H8 = 8.0 g
36.0 g
%C =
x 100% = 81.8%
44.0 g
8.0 g
%H =
x 100% = 18.2%
44.0 g
Calculate the percent composition of the following compounds
Ethane – C2H6 (80.0% C, 20.0% H)
Sodium bisulfate – NaHSO4 (19.2 % Na, 0.83% H, 26.7% S, 53.3% O)
Ammonium chloride – NH4Cl (26.2% N, 7.5% H, 66.4% Cl)
Calculate the percent nitrogen in these common fertilizers
CO(NH2)2 (46.7% N)
NH3 (82.4% N)
NH4NO3 (35.0% N)
-
Using Percent as a Conversion Factor
o We can use % composition as a conversion factor in determining
the mass of one element in a given mass of a compound.
Example Problems
Calculate the mass of carbon in 82.0 g of propane (C3H8).
Known
Unknown
mass of carbon = ? g C
Mass of C3H8 = 82.0 g
First we must calculate the % composition of carbon in C3H8
36.0 g
x 100% = 81.8%
%C =
44.0 g
Next we use the % composition in relationship to 100%
mass of compound x
82.0 g C3 H 8 =
% of element in grams
= mass of element in compound
100% in grams
81.8 g C
= 67.1 g C
100.0 g C3 H 8
Calculate the mass of hydrogen in each of the following:
350 g C2H6 - 70 g H
20.2 g NaHSO4 – 0.17 g H
2.14 g NH4Cl – 0.16 g H
David V. Fansler – Beddingfield High School – Page 16
Chemistry Lesson #6 - Chemical Quantities
Calculate the grams of nitrogen in the 125 g of each fertilizer:
CO(NH2)2 – 58.4 g N
NH3 – 103g N
NH4NO3 – 43.8 g N
-
Calculating Empirical Formulas
o The empirical formula gives the lowest whole number ratio of
atoms in a compound – the empirical formula for H2O2 would be
HO – As seen, the empirical formula may or may not be the same
as the molecular formula.
o The empirical formula tells us about the kinds of atoms and the
relative count of atoms, or moles of atoms in a molecule or
formula unit
o As noted above, empirical formula can be interpreted at the
microscopic level (the atom) or the macroscopic level (the mole).
o If you use the mole, the empirical formula is the lowest whole
number ratio of moles that combine to form a compound
o Again, the empirical formula may or may not be the same as the
molecular formula – if is not, then the molecular formula is a
simple multiple of the empirical formula
o So how do you calculate an empirical formula – you use percent
composition!
Sample Problem
What is the empirical formula of a compound that is 25.9% nitrogen and 74.1%
oxygen?
David V. Fansler – Beddingfield High School – Page 17
Chemistry Lesson #6 - Chemical Quantities
Known
% of nitrogen = 25.9% N
% of oxygen = 74.1% O
Unknown
empirical formula = N?O?
Strategy:
The lowest whole number ration of moles of nitrogen atoms to oxygen atoms, or
empirical formula is to be calculated. The percent composition tells the ration of
masses of nitrogen atoms to oxygen atoms in the compound. The ratio of masses
is changed to a ratio of moles by using the conversion factors based on the molar
mass of each element. This mole ratio is then reduced to the lowest whole
number.
According to the percent composition, in 100.0 g of the compound there is 25.9 g
of nitrogen and 74.1 g of oxygen. These values are used to convert to moles:
1 mol element
mass of element x
= mol of element
gam of element
1 mol N
25.9 g N =
= 1.85 mol N
14.0 g N
1 mol O
74.1 g O =
= 4.63 mol O
16.0 g O
So the mole ratio of nitrogen to oxygen is N1.85O4.63 – which is not a lowest whole
number ratio. So we have a little more work to do. To get closer, we should
divide each of the values obtained by the smaller of the two values:
1.85 mol N
= 1 mol N
1.85
4.63 mol O
= 2.50 mol O
1.85
So we are closer, but we still do not have a lowest whole number ratio – to obtain
the lowest whole number ratio, we multiply both number by the smallest number
that will give a result as a whole number ratio – start with 2 and work you way
up!
1.00 mol N x 2 = 2.00 mol N = 2 mol N
2.50 mol O x 2 = 5.00 mol O = 5 mol O
So the empirical formula is N2O5.
Calculate the empirical formula of each compound
94.1% O, 5.9% H – OH
79.8% C, 20.2% H – CH3
67.6% Hg, 10.8% S, 21.6% O – HgSO4
27.59% C, 1.15% H, 16.09% N, 55.17% O – C2HNO3
1,6-diaminohexane is used to make nylon. What is the empirical formula given
62.1% C, 13.8% H, and 24.1% N? C3H8N
David V. Fansler – Beddingfield High School – Page 18
Chemistry Lesson #6 - Chemical Quantities
-
Calculating Molecular Formulas
Formula (name)
Classification of formula
Molar mass
CH
C2H2 (ethyne)
C6H6 (benzene)
CH2O (methanal)
C2H4O2 (ethanonic acid)
C6H12O6 (glucose)
Empirical
Molecular
Molecular
Emperical and molecular
Molecular
Molecular
13
26 (2 x 13)
78 (6 x 13)
30
60 (2 x 30)
180 (6 x 30)
-
From the above table we see that we have two groups of 3 compounds.
The first group contains carbon and hydrogen, the second group contains
carbon, hydrogen and oxygen
- Also note that each compound in a group has a number of atoms that is a
whole number ratio, thus a different molar mass
- Just as we could determine the empirical formula of a compound, we can
also determine the molecular formula of a compound if we know the
empirical formula and the molar mass of the molecular compound
o From the empirical formula, we can calculate the empirical
formula mass (efm) – just the molar mass of the empirical formula
o The known molar mass of the compound is then divided by the
empirical molar mass – this gives the number of empirical formula
units in a molecule of the compound, and is the multiplier to
convert the empirical formula to a molecular formula
Sample Problems
Calculate the molecular formula of the compound whose molar mass is 60.0 g and
has an empirical formula of CH4N.
Known
Unknown
Empirical formula = CH4N
molecular formula = ?
Molar mass = 60.0g
EFM = 12.0 + 4.0 + 14.0 = 30.0g
molar mass of compound 60.0 g
multiplier =
=
=2
efm
30.0 g
so 2 x C, 2 x H4, and 2 x N = C2H8N2
Find the molecular formula of each compound give its empirical formula and
molar mass
Ethylene glycol (CH3O), used in anti-freeze, molar mass = 62 g/mol – C2H6O2
p-dichlorobenzene (C3H2Cl) (mothballs), molar mass = 147 g/mol – C6H4Cl2
Which pairs of molecules have the same empirical formula?
a. C2H4O2, C6H12O6 - yes
b. NaCrO4, Na2Cr2O7 - no
David V. Fansler – Beddingfield High School – Page 19
Chemistry Lesson #6 - Chemical Quantities